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AREA AND VOLUME WHERE DO THEFORMULAS COME FROM?Roger YarnellJohn Carroll University, ryarnell16@jcu.edu

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Recommended CitationYarnell, Roger, "AREA AND VOLUME WHERE DO THE FORMULAS COME FROM?" (2016). Masters Essays. 33.http://collected.jcu.edu/mastersessays/33

AREA AND VOLUME

WHERE DO THE FORMULAS COME FROM?

An Essay Submitted to the

Office of Graduate Studies

College of Arts & Sciences of

John Carroll University

In Partial Fulfillment of the Requirements

For the Degree of

Masters of Arts in Mathematics

By

Roger L Yarnell

2016

This essay of Roger L. Yarnell is hereby accepted:

________________________________________________ __________________

Advisor – Douglas Norris Date

I certify that this is the original document

________________________________________________ __________________

Author – Roger L. Yarnell Date

1

TABLE OF CONTENTS

1. Introduction 2

2. Basic Area Formulas 4

3. Pi and the Circle 10

4. Basic Volume Formulas 14

5. Sphere 17

6. Pythagorean Theorem 20

7. Conclusion 23

8. References 24

2

1. INTRODUCTION

What are area and volume? This was a question that was posed to incoming

geometry students that have previously worked with these measurements in middle school.

There were a variety of responses to the question, how are area and volume defined?

Approximately half of the students had a general idea of what area and volume measure.

The students stated that area is the amount of space in a two dimensional figure and volume

is the amount of space in a three dimensional object, the students made no mention as to

the units used to measure area and volume. Only one student stated that area is measured

in square units and volume in cubic units. Other student responses for area, included

multiplying two sides (the length times the width) and the length around an object. Yet

other student responses for volume, included the width and length combined, the mass of

an object, the weight of an object, how much area is in an object, and it is found by

multiplying length, width and height together.

Why are there so many different misconceptions for area and volume? Part of the

reason for this may be that the students are given the formulas and are asked to simply

compute the area and volume of different shapes and objects by substituting in values and

plugging it into a calculator to get an answer. This does not give any fundamental

understanding as to what the students are calculating or give them the meaning as to how

this formula works for the selected object. As a result, students do not fully understand

what their answer represents nor does it have any real or true meaning.

The students need to know the meaning of area and volume and where the concepts

originate in order to give them meaning. When the students understand where concepts

come from they can obtain a deeper understanding and this will allow them to know the

meaning of their solution. The students have been shown why the area of a rectangle is

length times width but not for other shapes. This is why the students remember and think

area is simply length times width. The students use the other formulas not knowing why

those formulas work to find area for a particular figure.

As the instructor, I was not proving all the formulas for area and volume of objects.

I would prove some of the formulas, such as area of a rectangle, square, parallelogram,

3

triangle, and trapezoid and the volume of a rectangle prism. I thought these could be done

with complete understanding but not the other formulas with more complicated proofs.

Due to time constraints, the decision was made that it was more important for the students

to apply the formulas to problem solving. Problem solving is a very important skill, but

after further education in the graduate courses, it was revealed how important it is for the

students to understand where the formulas for area and volume originate and how exactly

the formulas give the desired result.

This is a unit that includes the history of area and volume and will demonstrate and

prove the formulas of each figure and shape that are used to calculate both area and volume.

This unit will have the student gain an understanding of the mathematics behind both area

and volume. This unit will also further enhance an appreciation and memory of what the

students are calculating and understand the meaningfulness of their answer.

What are area and volume? They are defined as follows [6]:

AREA is the number of unit square inside a closed region.

VOLUME is the number of unit cubes in a solid figure.

The earliest beginnings of geometry can be traced back to around 3000 BC to

ancient Egypt and Mesopotamia. They had used lengths, angles, areas, and volumes for

construction and surveying purposes. [5] The Babylonian and Egyptian people knew the

basic formulas for area and volume. They had descriptions for the area of several figures,

some correct and for others, such as the triangle, we are not sure they were completely

correct. They regarded geometry as a practical tool for their civilization. It is believed that

they did not have deductive proofs of their formulas. [7]

4

2. BASIC AREA FORMULAS

The area of a rectangle is probably the easiest to visualize. When the rectangle is

divided into unit squares the formula for area can be developed. By counting these unit

squares, the area of the rectangle can be determined. Also, the value for area can be found

by doing repeated addition. Count the

number of squares in each row, the number

of rows within the rectangle, and the

product will be the area. As in the diagram,

there are five squares in a row and this is to

be counted two times because there are two

rows. Repeated addition in this manner is

multiplication, so the number of squares

can be found by multiplying the dimensions

of length and width as this method of counting is a definition of multiplication. Therefore,

the area of a rectangle can be defined as length multiplied by width.

AREA OF A RECTANGLE: A = LW

The area of a square can also be found by applying the formula for the area of a

rectangle. A square is a rectangle with four congruent sides. If each side of the square is

represented by the variable S, using the formula A=lw and substituting S for both length

and width, then A = S∙S = S2 is obtained. Therefore the formula for the area of a square is

A = S2.

AREA OF A SQUARE: A = S2

The formula for the area of a parallelogram can be derived from the area of a

rectangle by taking the right triangle from one side of parallelogram and placing it onto the

other side to form a rectangle. As shown in the diagram on the next page of the

parallelogram ABED, the right triangle ABC can be removed and placed on the right side

5

forming right triangle DEF. The length BE

is the base of the parallelogram with AC

being its height. After moving triangle ABC

to the position of DEF a rectangle of equal

area is formed. Segment AC is equal to DF

which are the height and also the width of

the rectangle ADFC. The base BE is also

congruent to CF since BC and EF are congruent. BE = BC + CE, by the segment addition

postulate and BE = CE + EF by substitution. CF = CE + EF, so by substitution BE = CF.

The base of the parallelogram is equal to the width of the rectangle. Therefore, since the

area of a rectangle is A = LW and the base of the rectangle is equal to the length of the

rectangle and the height of the parallelogram is equal to the width of the rectangle the area

of a parallelogram is equal to base times height. A = bh.

AREA OF A PARALLELOGRAM: A = bh

The formula for the area of a triangle can be derived by drawing a diagonal in a

parallelogram. The opposite sides of

a parallelogram are congruent. Using

parallelogram ABCD, in the diagram

to the right, by drawing diagonal BD

we create two congruent triangles

ABD and CDB. The tringles are

congruent using the SSS postulate,

since the opposite sides of a

parallelogram are congruent and BD = BD by the reflexive property. Since the triangles

are congruent the area of one triangle is equal to half of the area of the parallelogram.

Therefore the area of a triangle is equal to half of the base times its height. A = ½bh.

AREA OF A TRIANGLE: A = ½bh

Below is an activity that the students can complete to prove the formulas for the

areas of a parallelogram and triangle by using grid paper, a ruler, and scissors.

C

BE

D

A

6

1. Draw two large parallelograms using the condition: If two line segments are

parallel and congruent then the quadrilateral formed is a parallelogram.

Make two of the sides horizontal.

2. Draw two altitudes in each parallelogram from a vertex to a horizontal side.

3. Label each horizontal side with a “b” and each height with an “h”. Place the

labels in the interior of each parallelogram.

4. Cut out each parallelogram.

5. Using the first parallelogram make one cut along a height. By rearranging

the pieces can you make another figure that you know the area of?

Figure:

Area:

Since the area of the parallelogram and your new figure are the same, what

can you conclude about the area of a parallelogram?

Area =

6. Using the second parallelogram. Make one cut to form two triangles.

7. What is true about the two triangles?

8. Compare the area of each triangle to the area of the parallelogram.

7

9. What can you conclude about the area of a triangle? What is the formula to

find the area of a triangle?

Area =

The next figure is that of a special triangle that has a formula of its own. This is an

equilateral triangle. An equilateral triangle is also equiangular. So not only does it have all

of its sides congruent, but so are all of its angles. Which means that the equilateral triangle

is a regular polygon. Since, all the

angles are congruent each angle is equal

to 60o. First, draw an altitude, from one

of the vertices, point B in our diagram to

side AC. This altitude also creates two

congruent right triangles, by using the

HL Theorem as BC = AB and BD = BD. Angles ABD and CBD are congruent because

they are corresponding parts in congruent triangles, also making them each 300. AC is

bisected by the altitude as well because AD and DC are corresponding parts. So,

AD = ½AC. We can use the Pythagorean Theorem to find the height, BD, of the triangle.

Let BC be s, then BC = AC = s. So, AD = ½s. By applying the Pythagorean Theorem to

triangle BDA, BD2 + (½s)2 = s2. By applying algebraic properties BD2 = s2 - ¼s2= ¾s2.

Finally by raising both sides to the power of ½, we obtain the equation BD = √3

2𝑠. Now,

using the formula for the area of a triangle we obtain the formula A= ½∙√3

2𝑠∙s which

simplifies to the area formula of an equilateral triangle A = √3

4𝑠2.

AREA OF AN EQUILATERAL TRIANGLE: A = √𝟑

𝟒𝒔𝟐.

The formula for the area of a trapezoid can be derived from the area of a triangle.

By drawing a diagonal, EG in the diagram, the area of the trapezoid is divided into two

8

triangles. The triangles have the same

height since EF and DG are parallel. Let

EF = b1 and DG = b2 and the height = h.

The area of ΔEFG = ½b1h and the area of

ΔDEG = ½b2h. The area of the trapezoid

is the sum of the areas of the two triangles,

thus A= ½b1h + ½b2h. Then by factoring out the greatest common factor of ½h we obtain

a final formula of A = ½h(b1 + b2).

The formula for a trapezoid could also be derived by using parallelograms. A

congruent trapezoid can be formed by rotating the original trapezoid 180o around the lower

left vertex (see the first figure), then translating

the new image up and to the right so that the two

sides match forming a trapezoid (see the second

figure). The area of the trapezoid is now formed

by two congruent trapezoids, so the area of the

parallelogram is equal to double the area of the

original trapezoid. The area of the parallelogram

is equal to h(b1 + b2). This is twice the area of the

trapezoid. Therefore, two times the area of the

trapezoid equals h(b1 + b2). So, by multiplying

both sides by ½ we obtain A = ½h(b1 + b2).

AREA OF A TRAPEZOID: A = ½h(b1 + b2)

The formula for the area of a regular polygon, with n sides, can be derived by using

triangles. When a regular polygon is inscribed in a circle, the radius of the circle is the

distance from the center of the polygon to one of its vertices. The polygon can then be

divided into congruent isosceles triangles by drawing a radius to each of its vertices. This

will create the same number of congruent isosceles triangles as there are sides in the

polygon. To find the area of one of the triangles formed, an altitude also called the apothem

of the polygon must be drawn which is also a median in the triangle. This can proved using

9

the fact that the triangles are congruent.

When the altitude is drawn it forms two right

triangles, the triangles are congruent by the

Hypotenuse-Leg Theorem. Then by

corresponding parts the altitude bisects both

the base and the vertex angle of the isosceles

triangle making the altitude a median in the

triangle as well. To find the area of the

triangle we must first find the vertex angle.

To do this, divide 360o by the number of

triangles formed. The apothem bisects the vertex angle. Let s = the length of the side of the

polygon. Then the length of the side of the right triangle that is also part of the side of the

polygon will be 1/2s. By using the trigonometric function tangent, the length of the apothem

can be determined, which is also the height of the triangle. To find the area of the triangle

the formula, 1/2bh can be used. The height is the apothem and the base is the length of the

side of the polygon. So, A= 1/2as. There are n triangles in the n-sided polygon, therefore

multiplying the area of the triangle by the number of triangles n gives the area of the regular

polygon is A = 1/2ans. The perimeter of the regular polygon is the number of sides times

length of a side, p = ns. By substituting this into the formula, the simplified formula for the

area of a regular polygon is A = 1/2ap.

AREA OF A REGULAR POLYGON: A = ½ap

10

3. PI AND THE CIRCLE

What is pi? Pi is a well-defined constant. It is well defined by similarity as all

circles are similar. Pi is defined as follows:

Pi is circumference divided by diameter.

Where did pi come from? The earliest known recording of pi was from about 1650

BC by Ahmes from Egypt. He claimed that if you take 8/9 of the twice the radius and square

it, the area of a circle with that diameter will be obtained. Using his statement, A = (8/92r)2,

with some algebraic manipulations he obtained that pi = 256/81 ≈ 3.16049. About 100 years

later the Babylonians and Hebrews used 3 for pi. [3] Pi was also mentioned and talked

about in the Bible in 2 Chronicles chapter 4 verse 2.

“He made the sea of the cast metal, circular in shape, measuring ten cubits from

rim to rim and five cubits high. It took the line of thirty cubits to measure around

it.” [10]

This states that the diameter is 10 cubits and the circumference is 30 cubits. Using the ratio

that pi is equal to circumference divided by diameter, it can be calculated that three was

used for pi.

Pi was left at that approximation for several years until about 430 BC. Anitphon

and Bryson of Heraclea attempted to find the area of a circle by exhaustion. They inscribed

polygons in a circle. They started with a hexagon and then doubled the number of sides to

a dodecagon. They continued doubling the number of sides in the polygon but they made

little progress. Then about 200 years later Archimedes used the some concept of exhaustion

but focused on perimeters instead of area. Using both inscribed and circumscribed 96 sided

polygons he was able to claim that pi was between 31/7 and 310/71. The average of these two

values is about 3.1419 which is less than 0.003 units from the true value of pi. [3]

More than 650 years after Anitphon and Bryson, Lui Hui of China used the same

method of exhaustion with area and inscribed polygons. He was able to approximate pi to

be 3.1416 using a 3,072 sided polygon. Tsu Ch’ungchih and his son Tsu Keng-chih were

11

able to use a 24,756 sided polygon. They achieved an approximation of 355/113 or 3.1415929

for pi. No one found a more accurate value for pi for more than 1000 years. [3]

Francais Viete used the Archimedean Method of exhaustion using perimeters to

approximate pi. In 1579, he used a 393,216 sided polygon to approximate pi accurately to

10 decimal places. His biggest step toward getting the value for pi came when he described

pi as an infinite product. He broke the polygons into triangles and used the ratio of the

perimeters between the regular polygon and the second polygon with twice the number of

sides and the cosine of the angle using the half angle formula.

𝟐

𝝅 =

√𝟐

𝟐 x

√𝟐+ √𝟐

𝟐 x

√𝟐+ √𝟐+ √𝟐

𝟐 x ∙∙∙ [3]

Then in 1655, John Wallis used an infinite product that was much more efficient

than Francais Viete’s. He used the area of ¼ circles and small rectangles to come up with

the product below

𝝅

𝟐 =

𝟐

𝟏 𝒙

𝟐

𝟑 𝒙

𝟒

𝟑 𝒙

𝟒

𝟓 𝒙

𝟔

𝟓 𝒙

𝟔

𝟕 𝒙

𝟖

𝟕𝒙 ∙∙∙ [3]

In 1671, James Gregory and Gottfried Wilhelm Leibenz were able to write a series

to approximate pi using the arctan formula. The series that Gregory discovered was

simplified by Liebeniz in 1674. Leibeniz’s series to approximate for pi uses the angle π/4.

His series is listed below:

𝛑

𝟒 ≈ 1 -

𝟏

𝟑 +

𝟏

𝟓 -

𝟏

𝟕 +

𝟏

𝟗 -

𝟏

𝟏𝟏 + ∙∙∙

Then, by multiplying both sides of the equation by four an approximation for pi is achieved.

π ≈ 4 – 𝟒

𝟑 +

𝟒

𝟓 –

𝟒

𝟕 +

𝟒

𝟗 –

𝟒

𝟏𝟏 + ∙∙∙

While using this formula in 1794, Georg Vego calculated pi to 140 digits. [3]

Now using computers pi has been calculated accurately to over 51.5 billion digits.

But, why is the 16th letter of the Greek alphabet, π, used to represent the value of

pi? In the early days, c or p were used to represent the ratio of pi. In 1706, William Jones

used π to represent pi for the first time in a book that he published. He however was not

12

very influential in the mathematical world. However, Leonhard Euler was very influential

and he started using the symbol in 1737. With Euler using the symbol, the other

mathematicians started using the symbol π as well for pi. By 1794 most all of the

mathematicians in Europe were using π to represent the value of pi. [3]

Before deriving the formula for the area of a circle, the students will be asked what

exactly is pi? What does it represent? How may the number pi have been originally

developed? The students will complete an activity using string, a pencil, a piece of paper,

scissors, and a ruler. The students can be taken to a parking lot to discover what pi

represents.

1. Tie a piece of string to the pencil, place a point on the paper to represent the

center of the circle and make a circle using the string as the radius.

2. Measure the length of the diameter.

3. Cut a piece of string to the exact length of the diameter.

4. Take the string and measure the circumference of the circle using the length

of the diameter as a unit.

5. State the number of diameters needed to complete the length of the

circumference of the circle.

6. Repeat this procedure several times with different sized circles.

7. What can you conclude about pi and the number of diameters needed to

make the circle? Did the size of the circle effect the number of diameters?

13

Lastly, the formula for the area of a

circle will be derived. To derive the formula for

the area of a circle, divide a circle into

congruent sectors. Then arrange the sectors as

shown in the diagram below to approximately

form a rectangle. As the number of sectors

grows the object becomes more like a rectangle.

This will introduce, in a very basic sense, the

concept of a limit to the students. When the limit

of the number of rectangles approaches infinity

the shape of the sectors approaches a rectangle.

The width of the rectangle will approach the radius and the length will approach half of the

circumference. It is half of the circumference because half of the sectors are creating the

length of the rectangle. To find the area of the rectangle we use the formula A = lw. So, the

formula of A = ½Cr will be obtained. Circumference C is defined as 2πr. By substituting

this into the area formula, the formula A = ½∙2πr∙r is achieved which simplifies to A = πr2.

The formula for the area of a circle can also be obtained by using the limit process

by using an inscribed polygon. By allowing the number of sides of the inscribed polygon

to approach infinity it will become very close to a circle. By applying the formula for a

regular polygon A = ½ap, the perimeter will approach the circumference of the circle. The

apothem will also approach the radius in the limit process. By substituting C for p and r for

a the formula A = ½rC is obtained. As before, this gives A = πr2.

AREA OF A CIRCLE: A = πr2

14

4. THE BASIC VOLUME FORMULAS

The volume of a rectangular prism is probably the easiest to visualize. A rectangular

prism can be formed by stacking cubical unit

blocks as shown in the figure. By counting the

unit blocks in the figure, the volume of the

rectangular prism will be found. The number

of unit blocks covering the base of the prism

can be calculated by multiplying the length of

the base by its width since the base is a

rectangle. The unit blocks on the base are then stacked three times in our figure to form the

prism. So the volume, the number of unit blocks, can be found by multiplying the area of

the base by the height of the prism. Seeing that all prisms are built by stacking layers of

bases, the volume of any prism can be found by using the formula V = Bh, where B is the

area of the base. Since the base is a rectangle its area can be calculated using the formula

for the area of a rectangle, A = lw, therefore the volume for a rectangular prism is V = lwh.

VOLUME OF A RECTANGULAR PRISM: V = LWH

The students can complete the following activity to develop the formula for the volume

of a rectangular prism by using unit cubes as described above.

1. Build the following rectangular prisms and count the number of blocks

needed to complete each prism.

2. Complete the table below:

L W H Number of Blocks

3 2 2

4 3 1

5 2 2

15

3. Using the length, width and height of each prism, how can you calculate the

number of blocks necessary to build the prism?

4. Build another prism and check to see if your conjecture works.

5. Explain why you believe that this may work for any rectangular prism.

The next figure that a formula for volume can be derived is a special rectangular

prism, the cube. The formula for the volume of a cube can be derived from the volume of

the rectangular prism. The cube is a rectangular prism with all edges being congruent. Let

each edge be s, then length, width, and height would all be equal to s. By substituting s into

the formula for the volume of a rectangular prism V = lwh, the formula V = s∙s∙s is achieved,

which simplifies to V = s3.

VOLUME OF CUBE: V = s3

The formula for the volume of a cylinder can be derived quickly because a cylinder

can be thought of as a prism with a circular base. A cylinder can be made of stacked

congruent circles as a prism is made of stacked congruent polygons. The volume of a prism

is V = Bh, since the base is a circle then the base area B is the area of a circle, A = πr2. By

substitution the volume of a cylinder can be found by using the formula V = πr2h.

VOLUME OF CYLINDER: V = πr2h

The volume of a cone can be found by using calculus and rotation around the

x-axis to form a solid cone. Geometry students though have yet to take calculus and would

not understand the integration necessary to prove the volume of a cone using this process.

16

So, another process is needed to derive the

formula not using calculus. A cone can be thought

of as a pyramid with a circular base, so if the

formula for the volume of a pyramid can be

derived then the formula for the cone can be

derived. The volume of a pyramid can be derived

using calculus and cross sections which geometry

students will not understand. So, another method

to derive the formula without calculus is needed.

To derive the volume of a pyramid, without calculus, notice that six pyramids can be

inscribed in a cube with each face of the cube being the base of a pyramid and the vertex

of each pyramid being at the center of the cube. The height of each pyramid will be half

the length of a side of the cube. All six pyramids will then be congruent to each other since

their bases and heights are the same. Let each side of the cube be 2x, as shown in the

diagram, the height of each pyramid will be x and the base of each pyramid will be 2x . 2x.

Thus, the area of the base of each pyramid will be 4x2. The volume of the cube is then

V = (2x)3 = 8x3. Since all the pyramids are congruent and are inscribed in the cube the

volume of each pyramid is then 1/6 the volume of the cube. So, the volume of each pyramid

is 1/6(8x3). The formula can be rewritten as follows V = 1/6 ∙ 2 ∙ 4x2 ∙ x. Since the height of

the pyramid h is equal to x and the base area B of the pyramid is 4x2, by substitution the

formula V = 1/6 ∙ 2 ∙ B ∙ h is obtained. After simplifying the formula for the volume of a

pyramid is V = 1/3Bh.

VOLUME OF A PYRAMID: V = 1/3Bh

The volume formula for a cone can now be derived using the formula for a pyramid.

The cone has a circular base which has an A = πr2. So, by substituting this for base area in

the volume formula for a pyramid the formula for the volume of a cone is V = 1/3πr2h.

VOLUME OF A CONE: V = 1/3πr2h

17

5. THE SPHERE

The final formula for volume that will be derived is for the sphere. Again the

volume formula for a sphere can be derived using calculus and rotating a semicircle around

the x-axis. In pre-calculus and geometry courses the formula needs to be derived without

integration Cavalieri’s Principle be used to derive the formula. The principle states that if

the areas of the cross sections of two solids are equal and the height of the two solids are

equal then the volumes of the two solids are equal. [4]

To derive the formula for the volume of a sphere two objects will be given. The

first is a sphere with a radius of r

inscribed in a cylinder and two inverted

cones each with their radii and heights

being equal to r inscribed in a cylinder.

First, take a cross sectional slice across

the sphere and the cones inscribed in the

cylinders that is h units from the center.

The cross section of the sphere will be a

circle, as shown in the figure below. The radius of the cross

section of the sphere is calculated using the Pythagorean

Theorem. One leg of the right triangle is h, the other is the

radius of the cross section, and the hypotenuse is r. Using the

Pythagorean Theorem the radius of the cross section will be

√𝑟2 − ℎ². So, the area of the cross sectional circle is

A = π(√𝑟2 − ℎ²)2 which simplifies to A = π(r2 – h2). The

cross section of the cone inscribed in the cylinder is a washer, as shown in the figure on

the next page. By looking at the front view of the two cones in the cylinder it will be noticed

that the cones form right isosceles triangles, the radius and the height both have length r.

By taking the cross section, h units from the center of the cylinder, the radius of the

cross sections of the cone will be h and the radius of the cylinder will be r. The vertical

r

h r

18

cross section through the cylinder and cones is a square. The horizontal cross sections have

the shape of a washer or annulus

with the cones corresponding to

the inner circle and the cylinder

corresponding to the outer circle.

So, the area of the horizontal

cross section will be A = πr2 –

πh2. By factoring out the greatest

common factor of pi the formula A = π(r2 – h2) will be achieved for area. This is the same

area as the cross section of the sphere also taken h units from the center. So, when all the

cross sections of each figure are combined, the volume of the sphere will be equal to the

volume of the cylinder minus the volume of the two cones. Using the volume formulas for

the cylinder and cone derived earlier, a formula for the volume of a sphere will be obtained

as follows, V = πr2h – 2(1/3πr2h) where h is the height of the cones. The height of the

cylinder is 2r and the height of the cone is r. By substituting these values for h into the

formula we obtain the formula V = πr2(2r) – 2(1/3πr2(r)) which then simplifies to V = 2πr3

– 2/3πr3. Thus, V = 4/3πr3 is the volume formula for a sphere.

VOLUME OF SPHERE: V = 4/3πr3

Next, the formula for the surface area of a sphere will be derived. This can be completed

now that the formula for the volumes of a sphere and pyramid have been discovered. The sphere

will be filled with pyramids in order to approximate its surface area. The vertex of each pyramid

will be at the center of the sphere. The volume of the

pyramid is equal to 1/3Bh, where B is the area of the base

of the pyramid. The volume of the sphere, can be

realized by approximating the interior of the sphere by

packing an infinite number of pyramids. The volume of

the sphere can be realized as the limit as the number of

pyramids approaches infinity of the summation of the

volumes of the pyramids. The height of each pyramid

will approach the radius of the sphere. Using this

process the volume of the sphere will be 1/3Ar, where A is the limit of the sum of the bases of all

the pyramids which is the surface area of the sphere. This formula is equal to the formula that was

h

r

r

h

h

r

19

derived earlier for the volume of a sphere giving the equation 4/3πr3 = 1/3Ar. By multiplying both

sides of the equation by 3 and dividing both sides by r the formula for the surface area of the sphere,

A = 4πr2 is derived.

SURFACE AREA OF SPHERE: 4πr2

20

6. THE PYTHAGOREAN THEOREM

There is another formula that is very important in geometry. It is used in solving

many problems that involve area and volume. This formula is the Pythagorean Theorem.

It is stated as follows: [8]

The Pythagorean Theorem states that in a right triangle the square of the

length of the hypotenuse is equal to the sum of the squares of the lengths of the

legs.

Hypotenuse2 = Leg12 + Leg2

2

Given the right triangle below, the Pythagorean Theorem is also quite often stated

following the definition using the triangle below obtaining the formula below:

c2 = a2 + b2

The Pythagorean relationship was known by the Babylonians about 1600BC. [1]

At this time the relationship had yet to be proved. Pythagoras, who lived from about

569 BC to 500 BC, imported proof into mathematics. This was probably his greatest

achievement. [2] He provided

a proof of the Pythagorean

Theorem. His proof consisted

of the rearrangement of

triangles as shown in the

diagram to the right. The two

large squares are identical and

the four triangles are all

21

congruent. The only difference is the white regions, thus giving the result of c2 = a2 + b2.

[9]

This is just one proof of this famous theorem. Another way to prove The

Pythagorean Theorem involves algebra. It uses the left side of the diagram on the previous

page. The length and width of the large square are a + b. So, the area of the large square is

(a + b)2 = a2 + ab + b2. The area of the square can also be found by adding the area of the

four triangles to the area of the small square. The four triangles are all congruent by the

SAS postulate. The area of each triangle is ½ab, so the area of all four triangles combined

is 4(½ab) = 2ab. The area of the small white square is c2, since each side is c. So the area

of the square obtained by adding all the triangles and the small square is 2ab + c2. By taking

the two calculations for the area of the square and setting them equal to each other the

equation a2 + 2ab + b2 = 2ab + c2 is obtained. Subtract 2ab from both sides and the

Pythagorean Theorem a2 + b2 = c2 is achieved with c representing the length of the

hypotenuse and with a and b representing the lengths of the legs of the right triangles.

The Pythagorean Theorem was also proved by United States President James A.

Garfield, then a United States Representative. He

proved the Pythagorean Theorem using a trapezoid

instead of squares. [9] He found the area of the entire

trapezoid using the formula for the trapezoid. The

height is a + b and one base is a while the other is b.

By using the formula for the area of a trapezoid the

equation A = ½(a + b) (a +b) is obtained. This is

equivalent to A = ½(a2 + 2ab + b2). Next, the area of

the trapezoid can be found combining the area of its

parts. Triangles 1 and 2 are congruent by the SAS

postulate. The area of each of those triangles is ½ab

while the area of the third triangle is ½c2. By adding the results of all three triangles the

formula A = ab + ½c2 is obtained. Then take the two results and set them equal to each

other obtaining the equation ½(a2 + 2ab + b2) = ab + ½c2. Multiply both sides by 2, the

22

result is a2 + 2ab + b2 = 2ab + c2. Then subtract 2ab from both sides and the result is the

conclusion of the Pythagorean Theorem a2 + b2 = c2.

A forth way to prove the Pythagorean Theorem is by using similar triangles. In the

diagram below there are three similar triangles, ΔABC ~ ΔACD ~ ΔCBD all by the AA

postulate. Two of the similar triangles will be

used at a time. First, using ΔABC ~ ΔACD, two

of the sets of proportional sides give the

equation c/b = b/x. By multiplying both sides by

bx, the equation cx = b2 is achieved. Second,

using ΔABC ~ ΔCBD, two of the sets of

proportional sides give the equation c/a = a/y. By

multiplying both sides by ay, the resulting equation is cy = a2. Next, add the two equations

together obtaining cx + cy = a2 + b2. This is equivalent to the equation c(x + y) = a2 + b2.

But, x + y = c so by substitution c ∙ c = a2 + b2 is the result and therefore c2 = a2 + b2.

There are many other ways to prove the Pythagorean Theorem. The previous four

are proofs that geometry students would be able to understand fully. By proving this

theorem the students will be able to understand where it comes from and why it works

when they use it in solving many different types of problems including area and volume

problems and applications.

23

7. CONCLUSION

Understanding the history and the origins of the concepts of area and volume and

their formulas is imperative to the comprehension and appreciation of the meaning of the

solutions the students achieve to applied problems. Demonstrating the proofs for the

formulas of each figure combined with the historical perspective will enhance student

learning. Students will be better able to recall and understand their calculations and what

they represent allowing them to be more proficient at problem solving. They will be able

understand the meaning of their solutions and will gain the ability to improve their

mastery of area and volume.

With an improved understanding of area and volume, the students will be more

successful and confident in their mathematical studies leading to more excitement about

mathematics. The students will be able to use these concepts later in their mathematical

studies to enhance their success in future mathematical courses. Students will be able to

visualize how these concepts are used thereby gaining the fundamental knowledge base

for selected coursework in later courses such as calculus. This confidence and

understanding will allow the students to have more success in college or career

endeavors.

24

8. REFERENCES

1. Babylonian Mathematics (2105, October 21). Retrieved November 13, 2015, from

Wikipedia: The Free Encyclopdia:

https://en.wikipedia.org/wiki/Babylonian_mathematics

2. Bell E.T., Men of Mathematics. New York: Simon and Schuster, Inc. (1937)

3. Blatner David, The Joy of π. New York: Walker Publishing Company (1997)

4. Cavalieri's Principle. 2010-01-09. All Math Words Encyclopedia. Life is a Story

Problem LLC. http://www.allmathwords.org/en/c/cavalierisprinciple.html.

5. Geometry. (2015, October 19). Retrieved October 25, 2015, from Wikipedia: The

Free Encyclopedia: https://en.wikipedia.org/wiki/Geometry

6. Haenisch, Seigfried. Geometry. Minnesota: AGS Publishing (2005)

7. Kline Morris. Mathematical Thought from Ancient to Modern Times. New York:

Oxford University Press (1972)

8. Larson, Tom, Boswell, Laurie, Stiff, Lee. Geometry Concepts and Skills. Illinois:

McDougal Littell (2005)

9. Pythagorean Theorem (2015, October 23). Retrieved November 13, 2015, from

Wikipedia: The Free Encyclopdia:

https://en.wikipedia.org/wiki/Pythagorean_theorem

10. The Holy Bible, New International Version. Michigan: Zondervan Bible

Publishers (1978)